Spring 2014 - Problem 4

Lp spaces

Consider a sequence $\set{a_n \mid n \geq 1} \subseteq \br{0, 1}$ . For $f \in C\p{\br{0,1}}$ , let us denote
$\phi\p{f} = \sum_{n=1}^\infty 2^{-n} f\p{a_n}.$
Prove that there is no $g \in L^1\p{\br{0,1}, \diff{x}}$ such that $\phi\p{f} = \int f\p{x} g\p{x} \,\diff{x}$ is true for all $f \in C\p{\br{0,1}}$ .
Each $g \in L^1\p{\br{0,1}, \diff{x}}$ defines a continuous functional $T_g$ on $L^\infty\p{\br{0,1}, \diff{x}}$ by
$T_g\p{f} = \int f\p{x} g\p{x} \,\diff{x}.$
Show that there are continuous functionals on $L^\infty\p{\br{0,1}}$ that are not of this form.

Suppose such a $g$ exists. Observe that $\chi_{\set{a_1}}$ is bounded and upper semicontinuous (it is a characteristic function on a closed set), so there exists a sequence of continuous functions $\set{f^{\p{N}}_n}_n$ which decrease to $\chi_{\set{a_1}}$ . Then by dominated convergence,
$0 = \int \chi_{\set{a_1}} g \,\diff{x} = \lim_{n\to\infty} \int f_ng \,\diff{x} = \lim_{n\to\infty} \phi\p{f^{\p{N}}_n} = \sum_{n=1}^\infty 2^{-n} \chi_{\set{a_1}}\p{a_n} = \frac{1}{2},$
which is impossible.
If this were not the case, then $\p{L^\infty\p{\br{0,1}}}^* \simeq L^1\p{\br{0,1}}$ . But $L^1\p{\br{0,1}}$ is separable, which would imply that $L^\infty\p{\br{0,1}}$ is also separable, which is impossible.